Trigonometric Indentities $\dfrac{1}{\cos x}+ \dfrac{2\cos x}{\cos 2x}$
If $x=\dfrac{\pi}{7}$ Prove that above expression is equal to $4$.
I worked a few steps and reached here: 
$\dfrac{(4\cos^2 x -1)}{\cos x  (2\cos^2 x-1)}$
Not able to proceed after this!!!
 A: For $x=\frac{\pi}{7}$ it follows
$$\cos (4x)=-\cos(3x)\tag{1}$$
Let $y=\cos x$, using two times the cosine double-angle formula we get
$$\cos (4x)=2\cos^2 (2x)-1=2(2y^2-1)^2-1=8y^4-8y^2+1\tag{2}$$
And, from the identity
$$\cos (3x)=4\cos^3x-3\cos x\qquad\text{i.e.}\qquad \cos(3x)=4y^3-3y\tag{3}$$
we get
\begin{align*}
&&8y^4-8y^2+1&=-4y^3+3y\\
\iff&&8y^4+4y^3-8y^2+3y+1&=0\\
\iff&&(y+1)(8y^3-4y^2-4y+1)&=0
\end{align*}
Since $y=\cos \frac{\pi}7\neq -1$ it follows from the last equation
$$8y^3-4y^2-4y+1=0\qquad\iff\qquad4y^2-1=8y^3-4y=4y(2y^2-1)\tag{4}$$
On the other hand, the given expression can be written as
\begin{align*}
\frac1{\cos x}+\frac{2\cos x}{\cos (2x)}&=\frac{4\cos^2 x-1}{(\cos x)(2\cos^2x -1)}\\
&=\frac{4y^2-1}{y(2y^2-1)}\\
&=\frac{4y(2y^2-1)}{y(2y^2-1)}\qquad\text{from }(4)\\
&=4
\end{align*}
A: $$F=\dfrac{\cos2x+2\cos^2x}{\cos x\cos2x}=\dfrac{3-4\sin^2x}{\cos x\cos2x}$$
If $\sin x\ne0,$
$$F=\dfrac{\sin x(3-4\sin^2x)}{\sin x\cos x\cos2x}=\frac{2\sin3x}{\sin2x\cos2x}=4\cdot\dfrac{\sin3x}{\sin4x}$$
Now $\sin4x=\sin3x$
if $4x=n\pi+(-1)^n3x$ where $n$ is any integer
If $n$ is odd $=2m+1,$(say), $7x=(2n+1)\pi, x=\dfrac{(2n+1)\pi}7$ where $n\equiv0,\pm1,\pm2,\pm3\pmod7$
But $n\equiv3\pmod7\implies\sin x=0$
So, $\sin x\ne0\implies n\equiv0,\pm1,\pm2,-3\pmod7$
If $n$ is even $=2m$(say), $x=2m\pi\implies\sin x=?$
